Question 239915
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That depends on whether you mean


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{2}\ -\ 4i]


Which simply becomes


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ 4i]


Or


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{2}{2-4i}]


Which requires that you rationalize the denominator by multiplying by 1 in the form of the conjugate of your complex denominator divided by itself.


The conjugate of a complex number *[tex \Large a\ +\ bi] is *[tex \Large a\ -\ bi].  So you need to multiply your expression by *[tex \Large \frac{2+4i}{2+4i}]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{2}{2-4i}\right)\left(\frac{2+4i}{2+4i}\right)]


Since the denominator is now the factors of the difference of two squares, the imaginary part will disappear.  I'll let you do your own arithmetic, but remember *[tex \Large i^2\ =\ -1]


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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